Prove using the Triangle Inequality

1. Sep 23, 2011

Chinnu

1. The problem statement, all variables and given/known data

Show that:

(|x+y|)/(1+|x+y|) ≤ ((|x|)/(1+|x|)) + ((|y|)/(1+|y|))

2. Relevant equations

You are given the triangle inequality:

|x+y| ≤ |x| + |y|

3. The attempt at a solution

(This is done from the result, as I haven't been able to find the starting point)

(|x+y|)/(1+|x+y|) ≤ (|x|(1+|y|)+|y|(1+|x|))/((1+|x|)(1+|y|))

(|x+y|)/(1+|x+y|) ≤ (|x|+2|x||y|+|y|)/(1+|x|+|y|+|x||y|)

This doesn't seem to go anywhere. I also tried flipping the whole thing to get:

(1+|x+y|)/(|x+y|)≤(1+|x|)/(|x|)+(1+|y|)/(|y|)

but this doesn't seem to lead anywhere either...

2. Sep 23, 2011

spamiam

As a first step, try starting by applying the triangle inequality to the numerator of the left-hand side. Next, what can you say about $1 + |x+y|$ and $1+|x|$? What does this imply about $\frac{1}{1+|x+y|}$ and $\frac{1}{1+|x|}$?

3. Sep 24, 2011

Chinnu

so, since 1+|x+y| $\geq$ 1+|x|,

$\frac{1}{1+|x+y|}$ $\leq$ $\frac{1}{1+|x|}$

Now,

$\frac{|x+y|}{1+|x+y|}$ $\leq$ $\frac{|x|}{1+|x+y|}$ + $\frac{|y|}{1+|x+y|}$

So,

$\frac{|x|}{1+|x+y|}$ + $\frac{|y|}{1+|x+y|}$ $\leq$ $\frac{|x|}{1+|x|}$ + $\frac{|y|}{1+|x+y|}$

Can a similar argument now simply be extended for 1+|y|?

4. Sep 24, 2011

kru_

You're on the right track. Prove the inequality is true for the denominator, then you can easily prove it is true for the numerator.

5. Sep 24, 2011

spamiam

Actually, this isn't true: take x=1 and y=-1 for a counterexample.

I just noticed a trick that makes this much easier:

$$\frac{a}{1+a} = \frac{1+a-1}{1+a} = \frac{1+a}{1+a} - \frac{1}{1+a} = 1 - \frac{1}{1+a}\; .$$

Try using this trick on the LHS. Then you can use the triangle inequality (which will give you $1+|x+y| \leq 1 + |x| +|y|$) to compare $-\frac{1}{1+|x+y|}$ and $-\frac{1}{1+|x|+|y|}$. Then you can use the first trick backwards, which should lead to the result. And hopefully I haven't made any mistakes with my inequalities!